Lp estimates for the bilinear Hilbert transform
Abstract
For the bilinear Hilbert transform given by:
we
announce the inequality ∥H
fg∥p3 ≤
K
p1
,p2∥f∥p1∥g∥p2,
provided 2 < p
1,
p
2 < ∞, 1/p
3
= 1/p
1 +
1/p
2 and 1 <
p
3 < 2.
We announce a partial resolution to long standing conjectures
concerning the operator known as the bilinear Hilbert transform,
defined as follows:
This operation is initially defined only for certain functions
f and g, for instance those in the Schwartz class
on ℝ. The conjectures concern the extension of H to a
bounded operator on L
p spaces. We have proved:
Theorem 1. H extends to a bounded operator on Lp1 × Lp2 into Lp3, provided 2 < p1, p2 < ∞ and 1 < p3 < 2, where 1/p3 = 1/p1 + 1/p2.
Some 30 years ago, in connection with the Cauchy integral on Lipschitz curves, Calderón (1) raised the question of H mapping L 2 × L 2 into L 1; this inequality is true. Indeed, the bilinear Hilbert transform maps into L p3 provided only that p 3 > 2/3.
Study of the bilinear Hilbert transform is intimately related to Carleson’s theorem (2) asserting the pointwise convergence of Fourier series. A seminal result, it has received two proofs, with the alternative proof provided by Fefferman (3). These proofs have provided us with ingenious and complementary methods of time frequency analysis. A similar analysis seems necessary to understand H, and so our proof entails significant aspects of both Carleson’s and Fefferman’s proofs. We give a description of our proof, with details presented in their most concrete form. Complete proofs, which appear in ref. 4, require definitions and constructions somewhat more general than those presented here.
The bilinear Hilbert transform must be broken into scales and the
frequency behavior of each scale understood. Hence we replace the
kernel 1/y with Σj=−∞
∞ 2jρ(2j
y), where ρ is a Schwartz
function with Fourier transform ρ̂(ξ) = ∫
e
−2πixξρ(x) dx supported on [1/2, 2). For
each j, consider:
which has bilinear symbol ρ̂(2−j(ξ −
θ)). More specifically,
Therefore, if f is supported in frequency on the
interval [n2j, (n + 1)2j],
then H
j
fg(x) acts on the inverse Fourier
transform of ĝ(ξ)1[(n +
1/2)2j, (n + 3)2j](ξ), and is
supported in frequency on the interval [(2n +
1/2)2j, (2n + 4)2j]. The differing
rates of translation make these three intervals distinct.
It is important to note that the location of the intervals is arbitrary, and therefore, for all j and j′, the inner product of H j fg and H j′ fg need not tend to zero as |j − j′| tends to infinity. The analysis of H must be done in terms of both time and frequency.
Instead of proceeding with a decomposition of H, we
define a model of it adapted to the combinatorics of the
time–frequency plane. Let 𝒟 be a dyadic grid in ℝ. Call
I × ω ∈ 𝒟 × 𝒟 a tile if |I|·|ω| =
1. The interval ω is a union of four dyadic subintervals of
equal length, ω1, ω2, ω3, and
ω4, which we list in ascending order. Thus,
ξi < ξj for all 1 ≤
i < j ≤ 4 and ξj ∈
ωj. (We will only use ωj
for j = 1, 2, 3.) We adopt the notation t =
I
t × ωt and tj =
I
t × ωtj for j = 1, 2,
3. Fix a Schwartz function φ with φ̂ supported on
[−1/8, 1/8], in addition require that ∫ φ(x −
16n)φ(x) dx = 0 for all integers n. Set for all
tiles t and j = 1, 2, 3,
where c(J) denotes the center of the interval J.
Then our model of the bilinear Hilbert transform is
which is initially defined only for Schwartz functions
f
1 and f
2. We emphasize
that the sum extends over all tiles, and hence all scales. The analogue
of Theorem 1 is
Theorem 2. ℳ extends to a bounded operator on Lp1 × Lp2 into Lp′3, provided 2 < p1, p2 < ∞ and 1 < p′3 = (1/p1 + 1/p2)−1 < 2.
With more liberal notions of “grid,” “tile,” and “φtj,” the bilinear Hilbert transform is in the convex hull of terms like our model ℳ.
In the present situation we can give a proof by way of duality. Thus we
take f
3 ∈
L
p3 and show that:
The sum is over positive quantities; namely, the decomposition
above already captures all of the cancellation necessary for
convergence of the sums. It also shows that the sum defining ℳ is
unconditionally convergent in t. And as each
f
j ∈
L
pj, where
p
j > 2, it follows that each function is
locally square integrable. As it turns out, L
2
arguments are decisive in proving Theorem 2.
We localize the sum above in the x variable by setting:
Certainly ∫ F
t(x) dx =
|I
t|−1/2 ∏j=1
3 |〈f
j, φtj〉|. And so we show that
F(x) = Σt
F
t(x) is
integrable. This follows from a weak-type result: for
p
j as above, there is a δ > 0 so that for all
|r
j − p
j| < δ, 1 ≤ j
≤ 3, the operator F(x) maps
L
r1 ×
L
r2 ×
L
r3 into
L
r,∞, where 1/r =
1/r
1 + 1/r
2 + 1/r
3. Then
a variant of the Marcinkiewicz interpolation theorem due to Janson (5)
implies the strong-type inequality.
A single instance of the weak-type inequality is:
for some constant K. But this inequality implies the
weak-type result, because F commutes with dilations by
powers of 2, and so it suffices to establish this last inequality.
These observations are useful since some of our estimates begin to
break down on exceptional sets of small measure. Due the localization
of F
t in the time variable and the fact that we
only aim for a distributional inequality, we can delete tiles
t whose time coordinate falls in a set of bounded measure.
The combinatorics of the time frequency plane enter in by way of the partial order on the tiles given by t < t′ if I t ⊂ I t′ and ω ⊃ ω′. Note that t and t′ are not comparable with respect to < if and only if t ∩ t′ = ∅. Being disjoint suggests orthogonality for the functions φtj and φt′j′, the dominant theme of Lemmas 1–3 we state below.
Call a collection of tiles T a Carleson–Fefferman (CF) set
with top q if t < q for all t
∈ T. Thus ωq ∩ ωt ≠
∅ for t ∈ T. Call T a
j-CF set if T is a CF-set for which the intervals
ωtj intersect for all t ∈
T. Notice that if T is a 1-CF set, say, then the
intervals {ωtj|t ∈ T} are
pairwise disjoint for j = 2, 3. Therefore, by
application of Cauchy–Schwartz:
Notice that the last two square functions are Littlewood–Paley
g functions, albeit conjugated by an exponential to account
for the location of the CF set in frequency.
This estimate forms the motivation for Lemma 1 below, which formalizes a decomposition of the set of tiles that is fundamental to our argument.
Lemma 1. Fix pi > 2. There is a δ > 0 and an ɛ0 > 0 and a constant K so that for all |ri −
pi| < δ and 0 < ɛ <
ɛ0, the following holds. The collection of all
tiles
S
is a union:
with these properties. First,
S
0
is trivial in that:
Then
S
n,i,j
is a union of
disjoint i-CF sets
T
q
with tops
q ∈ S
*
n,i,j, and:
Here, recall that 1/r = Σi
1/ri, which can be taken arbitrarily close to
1. And, most significantly, for t =
min
i{pi/2} − ɛ,
With Lemma 1 in place, we estimate:
The last sum is finite as r is arbitrarily close to
one, while t + ɛ = min{p
i/2}
> 1 is a fixed distance from one. Therefore, with Eq.
3, Eq. 1 holds.
We cannot give the complete construction of the
S
n,i,j, but rather the initial steps, in
which the nearly orthogonal classes of φti are
identified. First, we make an important comparison to a maximal
function. If T
q is an i-CF set
with top q, we have for j ≠ i,
Here M
2
g is the maximal function
(M|g|2)1/2. Thus the set
F = ∪i{M
2
f
i >
C
−1} has bounded measure and we define
S
0 = {s|I
s ⊂ F},
making Eq. 4 trivial. For all i-CF sets
T with top q, and T ⊂
S/S
0, we have Δ(T, j) ≤ 1,
for j ≠ i.
The remaining construction is inductive. Assume that the
S
m,i,j are defined for all m <
n and all i, j, in such a way that for
S
r =
S/∪m<n∪i,j
S
m,i,j)
we have:
and for any i-CF set T
q ⊂
S
r with top q,
Δ(T
q, j) ≤
2−n/rj
+2,
for j ≠ i. As the same inequality applies to each
sub-CF set of T, we conclude that:
We define S*
n,1,1 to be the
set of maximal tiles q with |〈f
1,
φq,1〉| ≥
2−n/r1−1
,
and take S
n,1,1 to consist of all tiles
t so that t1 < q for some q
∈ S*
n,1,1. These tiles are
removed, and then S
n,i,i is defined
similarly for i = 2, 3. After the deletion of the
tiles D
0 =
∪i=1
3
S
n,i,i, we have
|〈f
i, φti〉| ≤
2−n/ri
−1
for all tiles t ∈ S
r′ =
S
r/D
0.
The set S n,1,2 has a slightly different construction. Consider 1-CF sets T q ⊂ S r′ with top q so that Δ(T, 2) ≥ 2−n/r2 +1. We take T q to be the maximal 1-CF set with this property. Let q(1) be such a top, which is maximal with respect to <, and in addition sup{ξ|ξ ∈ ωq} is maximal. Remove the tiles T q(1), and repeat this procedure to define T q(2) and so on. S n,1,2 is then ∪ℓ T q(ℓ) and S* n,1,2 = {q(ℓ)|ℓ ≥ 1}. Observe that for any 1-CF set T ⊂ S r′/S n,1,2, we have Δ(T, 2) ≤ 2−n/r2 +1. These procedures are repeated inductively to define the S n,i,j for all n, i, j.
With the construction above it is elementary to check that these
properties hold.
And in the case of i ≠ j, the collection
S
n,i,j is a union of disjoint
i-CF sets T
q, with q
∈ S*
n,i,j, for which:
These last two bounds differ by a factor of 4, which is relevant
below. See the comments concerning the minimal tiles immediately
following Lemma 3 below. To achieve Eq. 4, one
must delete some tiles t, using Eq. 2, the upper
bounds Eqs. 6 and 7, and the control on the
number of trees given in Eq. 5.
The essence of the matter lies in the control of the number of CF-sets, which is in the verification of Eq. 5. Eq. 5 relies upon the inequalities in the previous paragraph and Lemma 2 and 3 below, which address the issue of almost orthogonality.
Let us consider S
n,1,1, say. The tiles
S*
n,1,1 are maximal and therefore
pairwise disjoint, which suggest weak orthogonality for the collection
of functions {φq,1|q ∈
S*
n,1,1}. If they were in fact
orthogonal, Bessel’s inequality and Eq. 8 implies:
While f
1 is not in
L
2, this inequality can be strengthened to an
analogous form for L
r for r > 2.
However, disjointness of tiles does not imply orthogonality, because
the functions φq,1 are not compactly supported
in the x variable. Indeed, by our choice of φ, for two
tiles t and s we have 〈φti,
φsi〉 = 0 if ωti ∩
ωsi = ∅ or ωti =
ωsi. But if ωti
⊂≠ ωsi then for all n ≥
0:
If we assume that a stronger separation of the tiles in the
x variable, then we would expect orthogonality. And in this
direction we have:
Lemma 2. For n ≥ 1 there are constants K and Kn
so that
the following holds for all A ≥ 1. Let
S
be any collection of tiles so that:
Here, for an interval I, AI denotes the interval
with the same center as I and length A|I|. Set NS(x) =
Σt∈S
1
It(x). Then:
A further combinatorial lemma asserts that if the tiles
{ti|t ∈ S′} are merely disjoint, then
after deleting tiles t for which I
t
falls in an exceptional set of small measure, S′ is a union
of O(A
3) collections of tiles S that
satisfy the stronger disjointness condition (Eq. 11).
The previous lemma is essential in obtaining Eq. 5 for the classes S n,i,i. A corresponding lemma is necessary for the S n,i,j, with i ≠ j, with Eq. 9 replacing the role of Eq. 8. It is:
Lemma 3. For n ≥ 1 there are
constants K and Kn
so that the
following holds for all A ≥ 1. Let
S
be a
union of j-CF sets
T
q
with tops
q ∈ S*
. Suppose that:
and for t ∈ T
q, q
∈ S*
and i ≠ j fixed,
Set NS(x) =
Σq∈S*
1
Iq(x). Then:
Notice that in a j-CF set
T
q, the tiles
{ωti|t ∈
T
q} are pairwise disjoint. Thus Eq.
12 is stronger than merely asserting that the tiles
{ωti|t ∈ T} are pairwise
disjoint. With the construction of the
S
n,i,j for i ≠ j given
above, Eq. 12 is true after deleting the minimal tiles
S
n,i,j
min in
S
n,i,j. The minimal tiles are controlled
with the first half of Eq. 9 and the observation that
Σs∈Sn,i,jmin
1
Is(x) ≤
N
n,i,j(x) for all x.
The method of proof of both Lemmas 2 and 3 is
similar. For instance, in Lemma 2, one considers the
operator:
If S is finite, this is a compact self-adjoint
operator, with maximal eigenvalue B. It suffices to estimate
B, as for all f ∈ L
2,
Σt∈S |〈f,
φti〉|2 = 〈f,
𝒮S
f〉 ≤ B∥f∥2
2.
Consider a normalized extremal eigenfunction f of
𝒮S. One then estimates:
which is expanded in diagonal and off-diagonal terms. The diagonal
term is Σt∈S |〈f,
φti〉|2 = 〈f,
𝒮S
f〉 ≤ B, which is an adequate
estimate for B
2. The off-diagonal term is by
Cauchy–Schwarz,
The innermost sum is bounded by
C
n
A
−n
infx∈ItMMf(x). This is seen by
invoking the estimate |〈f, φsi〉| ≤
K
infx∈Is
Mf(x), using Eq.
10 and carefully exploiting the geometry of the tiles via
the assumption 11.
One then sees that the off-diagonal term is no more than:
This, with the diagonal estimate, proves that B
2
≤ B +
C
n
B
1/2
A
−n∥N
S∥∞,
whence follows Lemma 2.
The research outlined herein is the product of several years of effort (6, 7).
Acknowledgments
The success of our efforts is due to the guidance and encouragement we have received from R. Coifman. M.L. has been supported by the National Science Foundation. M.L. and C.T. acknowledge the support of a North Atlantic Treaty Organization travel grant.
Footnotes
-
Charles Fefferman, Princeton University, Princeton, NJ
-
Abbreviation: CF, Carleson–Fefferman.
- Copyright © 1997, The National Academy of Sciences of the USA





